We have, $\sum_{k=1}^6\left[\sin \frac{2 k \pi}{7}-i \cos \frac{2 \pi k}{7}\right]$
$\begin{aligned}
& =\sum_{k=1}^6\left[(-i)\left(\cos \frac{2 k \pi}{7}+i \sin \frac{2 \pi k}{7}\right)\right] \\
& =(-i) \sum_{k=1}^6\left(\cos \frac{2 k \pi}{7}+i \sin \frac{2 \pi k}{7}\right)=(-i) \sum_{k=1}^6 \alpha^k
\end{aligned}$
Let $\alpha=\cos \frac{2 \pi k}{7}+i \sin \frac{2 \pi k}{7}$
$=(-i)\left(\alpha+\alpha^2+\alpha^3+\ldots+\alpha^5\right)$
Here, $\alpha+\alpha^2+\alpha^3+\ldots+\alpha^6$ follows G.P. so its sum.
$\mathrm{S}=\frac{\alpha\left(1-\alpha^6\right)}{1-\alpha}=\frac{\alpha-\alpha^7}{1-\alpha}=\frac{\alpha-1}{1-\alpha}=i$
$$
\left[\because \alpha^7=\cos 2 \pi+i \sin 2 \pi=1\right]
$$
Thus, $\alpha=-i(-1)$
$\alpha=i$